
The contribution from each \(n\)  is of order
\(q^{nN+\delta _n}\)  where \(\delta _n\ge 0\)  are
non-negative integers.
In the large \(N\)  limit the sectors with \(n\ge 1\)  decouples and
only the \(n=0\)  sector contributes to the index.
Then (REF ) reduces to \({\cal I}_{U(\infty )}={\cal I}_{\rm KK}\) ,
which was confirmed in [1]}.
